Math 151 - Exam 2A - solutions
Problem 1 Find
dy
dx
where y = 2x5 + 20x + x3/5 .
Solution.
3
None of the big Rules needed. f 0 (x) = 10x4 + 20 + x−2/5 .
5
Problem 2 Consider the function
f (x) = x3 − 2x2 + 5.
a) Find f 0 (2).
b) What is the slope-intercept equation of the tangent line to the graph of
f (x) at x = 2?
Solution.
a) First, we compute f 0 (x) using the rules for powers of x:
f 0 (x) = 3x2 − 4x
Then we substitute x = 2 to get f 0 (2) = 12 − 8 = 4.
b) Since f (2) = 5, we have
y = 4(x − 2) + 5.
which expands to the slope-intercept form y = 4x − 3.
Problem 3 Find the derivative y 0 of the function below. Name any of
Product Rule (PR), Quotient Rule (QR), and Chain Rule (CR) that you
use. DO NOT SIMPLIFY.
2
y = e3x −4 .
Solution.
d u
du
= 6x,
e = eu , and
We use the Chain Rule with u = 3x2 − 4, so
dx
du
y 0 = 6xe3x
2 −4
.
Problem 4 Find the derivative y 0 of the function below. Name any of
Product Rule (PR), Quotient Rule (QR), and Chain Rule (CR) that you
use. DO NOT SIMPLIFY.
y = 7x3 ln x.
Solution.
We use the Product Rule with u = 7x3 and v = ln x, so u0 = 21x2 , v 0 = 1/x,
and
7x3
.
y 0 = 21x2 ln x +
x
If we had been asked to simplify, we’d have written
y 0 = 21x2 ln x + 7x2 .
but we have not.
Problem 5 Suppose f (x) and g(x) are functions with values f (0) = 3,
g(0) = 5, f 0 (0) = 2 and g 0 (0) = 1. Find h0 (0) when h(x) = f (x)/g(x).
Solution.
Use the Quotient Rule,
f 0 (x)g(x) − f (x)g 0 (x)
d f (x)
=
.
dx g(x)
g(x)2
Now we substitute the given information to get the value
h0 (0) =
2·5−3·1
7
= .
2
5
25
2
Problem 6 Let f (x) = x3 + 15x
− 18x + 3.
2
Find all the x-values at which the graph of f has a horizontal tangent line.
Solution.
f 0 (x) = 3x2 + 15x − 18 = 3(x2 + 5x − 6) = 3(x + 6)(x − 1).
This is zero for x = −6 and x = 1, and these x-values are exactly where the
graph of f has a horizontal tangent line.
Problem 7 The graph of a function f (x) is shown below. All answers in
this question are integers.
a) Indicate all x-values where f (x) has a relative maximum.
b) Indicate all x-values where f (x) has a relative minimum..
c) List all relative maximum values of f (x).
d) List all relative minimum values of f (x).
e) List all intervals where this function is decreasing.
Solution.
a) x = −2, 1, 3. b) x = −1, 2. c) y = 2, 3, 4. d) y = 1, 2.
e) The function is decreasing in [−2, −1], [1, 2].
Problem 8 Consider the function
f (x) =
x+5
x2 − 9
a) Determine all critical numbers for this function.
b) Determine the intervals where this function is increasing.
c) Determine the intervals where this function is decreasing.
d) Find all numbers where this function has a local maximum.
e) Find all numbers where this function has a local minimum.
Solution.
First, we compute
f 0 (x) =
(x2 − 9) − 2x(x + 5)
x2 + 10x + 9
=
−
.
(x2 − 9)2
(x2 − 9)2
a) The derivative equals zero for x = −1 and x = −9. It is undefined for
x = ±3. All these together are the critical numbers.
b) Sampling gives f 0 (x) > 0 in (−9, −3)) and in (−3, −1)). These are the
intervals where f (x) is increasing.
c) Sampling gives f 0 (x) < 0 in (−∞, −9), in (−1, 3)), and in (3, ∞. These
are the intervals where f (x) is decreasing.
d) From part b) and c), x = −1 is the only number where f (x) has a local
maximum.
e) From part b) and c), x = −9 are the only numbers where f (x) has a local
minimum.
Problem 9 Assume that the unit price of manufacturing x refrigerators is
p(x) = 600 −
1 2
x
100
(a) What is the revenue R(x) from selling these x units?
(b) Find the marginal revenue R0 (x) at x = 50.
(c) Use your answer in part (b) to approximate the additional revenue from
selling 6 extra units in addition to the 50 units from part (b). The exact
answer will not give credit.
Solution.
1 3
x.
(a) R = xp = 600x −
100
2
(b) R0 (x) = 600 − 3x
, so R0 (50) = 525.
100
(c) R(56) ≈ R(50) + 6R0 (50) = R(20) + 3150, so 3150 dollars additional
revenue.
Problem 10
price of
Assume that x batches of XBlasters can be sold at a unit
p(x) = 4.5 − 0.002x
and the total cost of manufacturing x units is
C(x) = 100 + 0.5x.
(a) What is the total profit P (x) from selling x batches?
(b) What is the average profit P̄ (x)?
(c) For what value of x is the profit P (x) maximal? Verify that you really
have a maximum.
Solution.
(a) P (x) = xp(x) − C(x) = 4x − 0.002x2 − 100.
(b) P̄ (x) = 4 − 0.002x − 100/x.
(c) We need to solve P 0 (x) = 0. First,
P 0 (x) = 4 − 0.004x
which only has the solution x = 1000. Since P 0 (x) < 0 for x > 1000 and
P 0 (x) > 0 for x < 1000, this really is the only maximum.